怎么把string拆分
一个string类型的字符串“0|0|201312|0|0|20000|800;0|0|201312|0|0|21425|800;0|7|201312|0|0|21425|24|1050000:21425|800……”我只想要冒号后面的,冒号后面的都是“21425|800”这种类型,而且我不能确定冒号后面会存在几个,如果存在几个就分成几个string,这个怎么写啊
[解决办法]
#include <stdio.h>
char s[]="123 ab 4";
char *p;
int v,n,k;
void main() {
p=s;
while (1) {
k=sscanf(p,"%d%n",&v,&n);
printf("k,v,n=%d,%d,%d\n",k,v,n);
if (1==k) {
p+=n;
} else if (0==k) {
printf("skip char[%c]\n",p[0]);
p++;
} else {//EOF==k
break;
}
}
printf("End.\n");
}
//k,v,n=1,123,3
//k,v,n=0,123,3
//skip char[ ]
//k,v,n=0,123,3
//skip char[a]
//k,v,n=0,123,3
//skip char[b]
//k,v,n=1,4,2
//k,v,n=-1,4,2
//End.
#include <stdio.h>
#include <string.h>
char string[80];
char seps1[3];
char seps2[3];
char *token;
char *zzstrtok (
char *string,
const char *control1,//连续出现时视为中间夹空token
const char *control2 //连续出现时视为中间无空token
)
{
unsigned char *str;
const unsigned char *ctrl1 = control1;
const unsigned char *ctrl2 = control2;
unsigned char map1[32],map2[32];
static char *nextoken;
static char flag=0;
unsigned char c;
int L;
memset(map1,0,32);
memset(map2,0,32);
do {
map1[*ctrl1 >> 3]
[解决办法]
= (1 << (*ctrl1 & 7));
} while (*ctrl1++);
do {
map2[*ctrl2 >> 3]
[解决办法]
= (1 << (*ctrl2 & 7));
} while (*ctrl2++);
if (string) {
if (control2[0]) {
L=strlen(string);
while (1) {
c=string[L-1];
if (map2[c >> 3] & (1 << (c & 7))) {
L--;
string[L]=0;
} else break;
}
}
if (control1[0]) {
L=strlen(string);
c=string[L-1];
if (map1[c >> 3] & (1 << (c & 7))) {
string[L]=control1[0];
string[L+1]=0;
}
}
str=string;
}
else str=nextoken;
string=str;
while (1) {
if (0==flag) {
if (!*str) break;
if (map1[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
break;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
string++;
str++;
} else {
flag=1;
str++;
}
} else if (1==flag) {
if (!*str) break;
if (map1[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
flag=0;
break;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
*str=0;
str++;
flag=2;
break;
} else str++;
} else {//2==flag
if (!*str) return NULL;
if (map1[*str >> 3] & (1 << (*str & 7))) {
str++;
string=str;
flag=0;
} else if (map2[*str >> 3] & (1 << (*str & 7))) {
str++;
string=str;
} else {
string=str;
str++;
flag=1;
}
}
}
nextoken=str;
if (string==str) return NULL;
else return string;
}
void main()
{
strcpy(string,"A \tstring\t\tof ,,tokens\n\nand some more tokens, ");
strcpy(seps1,",\n");strcpy(seps2," \t");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234
[解决办法]
LIYI
[解决办法]
China
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK");
strcpy(seps1,"
[解决办法]
");strcpy(seps2," ");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234
[解决办法]
LIYI
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK");
strcpy(seps1,"");strcpy(seps2,"
[解决办法]
");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"1234
[解决办法]
LIYI
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK");
strcpy(seps1,"
[解决办法]
");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",a");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,"a,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",a,,b");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,a,,b,,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,");
strcpy(seps1,",");strcpy(seps2,"");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
strcpy(string,",,,");
strcpy(seps1,",");strcpy(seps2," ");
printf("\n[%s]\nTokens:\n",string);
token=zzstrtok(string,seps1,seps2);
while (token!=NULL) {
printf(" <%s>",token);
token=zzstrtok(NULL,seps1,seps2);
}
}
//
//[A string of ,,tokens
//
//and some more tokens,]
//Tokens:
// <A>, <string>, <of>, <>, <tokens>, <>, <and>, <some>, <more>, <tokens>, <>,
//[1234
[解决办法]
LIYI
[解决办法]
China
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK]
//Tokens:
// <1234>, <LIYI>, <China>, <010>, <201110260000>, <OK>,
//[1234
[解决办法]
LIYI
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK]
//Tokens:
// <1234>, <LIYI>, <010>, <201110260000>, <OK>,
//[1234
[解决办法]
LIYI
[解决办法]
010
[解决办法]
201110260000
[解决办法]
OK]
//Tokens:
// <1234>, <LIYI>, <>, <010>, <201110260000>, <OK>,
//[a]
//Tokens:
// <a>,
//[a,b]
//Tokens:
// <a>, <b>,
//[a,,b]
//Tokens:
// <a>, <>, <b>,
//[,a]
//Tokens:
// <>, <a>,
//[a,]
//Tokens:
// <a>, <>,
//[,a,,b]
//Tokens:
// <>, <a>, <>, <b>,
//[,,a,,b,,]
//Tokens:
// <>, <>, <a>, <>, <b>, <>, <>,
//[,]
//Tokens:
// <>, <>,
//[,,]
//Tokens:
// <>, <>, <>,
//[,,,]
//Tokens:
// <>, <>, <>, <>,
#include <stdio.h>
#include <string.h>
#define MAXN 100 //:后最多有
[解决办法]
间隔的MAXN项
#define MAXL 10 //每项最长10个ASCII字符
#define MAXLQ "10" //每项最长"10"个ASCII字符
char s1[]="0
[解决办法]
0
[解决办法]
201312
[解决办法]
0
[解决办法]
0
[解决办法]
20000
[解决办法]
800;0
[解决办法]
0
[解决办法]
201312
[解决办法]
0
[解决办法]
0
[解决办法]
21425
[解决办法]
800;0
[解决办法]
7
[解决办法]
201312
[解决办法]
0
[解决办法]
0
[解决办法]
21425
[解决办法]
24
[解决办法]
1050000:21425
[解决办法]
800";
char s2[]="0
------解决方案--------------------
0
[解决办法]
201312
[解决办法]
0
[解决办法]
0
[解决办法]
20000
[解决办法]
800;0
[解决办法]
0
[解决办法]
201312
[解决办法]
0
[解决办法]
0
[解决办法]
21425
[解决办法]
800;0
[解决办法]
7
[解决办法]
201312
[解决办法]
0
[解决办法]
0
[解决办法]
21425
[解决办法]
24
[解决办法]
1050000:21425
[解决办法]
800
[解决办法]
1
[解决办法]
9999999";
char t[MAXN][MAXL+1];
char *p,c;
int i,n,r;
int main() {
p=strchr(s1,':');
if (NULL==p) {
printf("[%s] not include :\n",s1);
return 1;
}
p++;
i=0;
while (1) {
r=sscanf(p,"%"MAXLQ"[^
[解决办法]
]%c%n",t[i],&c,&n);
if (2==r
[解决办法]
1==r) {
i++;
if (i>=MAXN) {
printf("Too many items!(>%d)\n",MAXN);
break;
}
if (1==r) break;
p+=n;
} else {//EOF==r
[解决办法]
0==r
break;
}
}
n=i;
for (i=0;i<n;i++) printf("t[%d]==[%s]\n",i,t[i]);
p=strchr(s2,':');
if (NULL==p) {
printf("[%s] not include :\n",s2);
return 1;
}
p++;
i=0;
while (1) {
r=sscanf(p,"%"MAXLQ"[^
[解决办法]
]%c%n",t[i],&c,&n);
if (2==r
[解决办法]
1==r) {
i++;
if (i>=MAXN) {
printf("Too many items!(>%d)\n",MAXN);
break;
}
if (1==r) break;
p+=n;
} else {//EOF==r
[解决办法]
0==r
break;
}
}
n=i;
for (i=0;i<n;i++) printf("t[%d]==[%s]\n",i,t[i]);
return 0;
}
//t[0]==[21425]
//t[1]==[800]
//t[0]==[21425]
//t[1]==[800]
//t[2]==[1]
//t[3]==[9999999]
//